我试图计算一种颜色（或最接近17种颜色之一）出现在图像像素中的次数（给定为300x300x3 np数组，浮点值[0,1]）。我已经写了这个，但是看起来效率很低：
for w in range(300):
for h in range(300):
colordistance = float('inf')
colorindex = 0
for c in range(17):
r1 = color[c, 0]
g1 = color[c, 1]
b1 = color[c, 2]
r2 = img[w, h, 0]
g2 = img[w, h, 1]
b2 = img[w, h, 2]
distance = math.sqrt(
((r2-r1)*0.3)**2 + ((g2-g1)*0.59)**2 + ((b2-b1)*0.11)**2)
if distance < colordistance:
colordistance = distance
colorindex = c
colorcounters[colorindex] = colorcounters[colorindex] + 1
有什么方法可以提高效率吗？我已经在对外部循环使用多处理了。
您提到您正在使用numpy，因此应避免尽可能地迭代。我的矢量化实现速度提高了约40倍。我对您的代码进行了一些更改，以便它们可以使用相同的数组，从而可以验证正确性。这可能会影响速度。
import numpy as np
import time
import math
num_images = 1
hw = 300 # height and width
nc = 17 # number of colors
img = np.random.random((num_images, hw, hw, 1, 3))
colors = np.random.random((1, 1, 1, nc, 3))
## NUMPY IMPLEMENTATION
t = time.time()
dist_sq = np.sum(((img - colors) * [0.3, 0.59, 0.11]) ** 2, axis=4) # calculate (distance * coefficients) ** 2
largest_color = np.argmin(dist_sq, axis=3) # find the minimum
color_counters = np.unique(largest_color, return_counts=True) # count
print(color_counters)
# should return an object like [[1, 2, 3, ... 17], [count1, count2, count3, ...]]
print("took {} s".format(time.time() - t))
## REFERENCE IMPLEMENTATION
t = time.time()
colorcounters = [0 for i in range(nc)]
for i in range(num_images):
for h in range(hw):
for w in range(hw):
colordistance = float('inf')
colorindex = 0
for c in range(nc):
r1 = colors[0, 0, 0, c, 0]
g1 = colors[0, 0, 0, c, 1]
b1 = colors[0, 0, 0, c, 2]
r2 = img[i, w, h, 0, 0]
g2 = img[i, w, h, 0, 1]
b2 = img[i, w, h, 0, 2]
# dist_sq
distance = math.sqrt(((r2-r1)*0.3)**2 + ((g2-g1)*0.59)**2 + ((b2-b1)*0.11)**2) # not using sqrt gives a 14% improvement
# largest_color
if distance < colordistance:
colordistance = distance
colorindex = c
# color_counters
colorcounters[colorindex] = colorcounters[colorindex] + 1
print(colorcounters)
print("took {} s".format(time.time() - t))
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